Using integers $a$ and $x$ in $a^x$, why does the # of digits of the exponentiated power multiplied by the $\log_{10}(a)$ equal $\pm1$ the exponent x?

Question

There have been multiple questions surrounding the determination of the number of digits in an integer power when given the base and exponent, here and here and here.

My question isn't how to determine how many digits are in the power produced by $a^x$ or how to compute the logarithm of $\log_a(B)$ where the logarithm, base and power are integers.

My question is why given $a,B$ where $a^x=B$, does counting the digits of $B$ base $10$ and multiplying the number of digits by the result of $\log_{10}(a)$ get us to within $1$ of $x$? Is there a particular logarithmic identity I'm missing?

There's some relationship I'm not connecting between $2^{56}$ = $10^{16.857679757182943}$ where the $\lfloor 16.857679757182943\rfloor+1$ is equal to the number of digits of both exponentiated powers $72057594037927936$ from $2^{56}$ and $10^{16.857679757182943}$.

Caveat: If based upon my naiveté in math, the preceding questions I referenced, should have given me my answer and therefore this question is duplicative I will remove it.

Answer 1

In your example, you have $$16 < \log_{10}{2^{56}} < 17,$$ which means that $$10^{16}<2^{56}<10^{17}.\tag{1}$$ Now, $10^{16}$ has $17$ digits, and $10^{17}$ is the smallest integer with $18$ digits, so $2^{56}$ has $17$ digits.

Taking base-two logarithms in $(1)$ gives $$16\log_2{10}<56<17\log_2{10}$$